Bounded invariant equivalence relations in elementary submodels. Let $\mathfrak{C}$ be the monster model of some theory $T$, let $E$ be a bounded invariant (over the empty set) equivalence relation on $\mathfrak{C}$ and let $\mathcal{M}\preccurlyeq \mathfrak{C}$ be any small elementary submodel. I want to prove that if two elements $a,b\in \mathfrak{C}$ have the same type over $\mathcal{M}$ then they are $E$ related.
What I have tried: There exists $\sigma\in Aut(\mathfrak{C}/\mathcal{M})$ such that $\sigma(a)=b$. If I can find $m\in \mathcal{M}$ such that $aEm$ then applying $\sigma$ we get $\sigma(a)Em$ and we are done. But I am not sure if I can find such an element $m$ for any arbitrary $a\in\mathfrak{C}$.
 A: The usual strategy is to first prove the following lemma.
Lemma: Let $E$ be a bounded invariant equivalence relation. If $I = (a_i)_{i\in \omega}$ is an indiscernible sequence, then $a_0E a_1$.
Proof: Suppose for contradiction that $a_0\not E a_1$. For any cardinal $\kappa$, we can extend the sequence $I$ to an indiscernible sequence of length $\kappa$. By invariance of $E$, no pair from the sequence is $E$-equivalent.  This shows that $E$ has at least $\kappa$-many equivalence classes, contradicting boundedness of $E$.
Ok, now suppose $M$ is a small model and $\text{tp}(a/M) = \text{tp}(b/M)$. Call this type $p(x)$. Let $p'(x)\in S(\mathfrak{C})$ be a global $M$-invariant extension of $p$, and let $(c_i)_{i\in \omega}$ be Morley sequence in $p'$ over $Mab$. Since $\text{tp}(a/M) = \text{tp}(b/M) = p'|_M$, the sequences  $a,c_0,c_1,\dots$ and $b,c_0,c_1,\dots$ are both Morley sequences in $p'$ over $M$. In particular, they are both indiscernible. So by the fact above, $aEc_0$ and $bEc_0$, and hence $aEb$.
